Puzzles with numbers and things

[davis_greatest]

d_g, can you clarify d(ii)

Does this mean that the first colour cannot be potted into the the black or pink pockets, i.e. unless (or until) a previous colour.....

Good question - I will amend the wording now to clarify it. What was intended is that, after potting a colour into any pocket except the yellow one, the pink and black pockets are out of bounds when potting the next colour. So, it is permissible to pot the first colour into the pink or black pocket.

[abextra]

looks quite high abextra are you sure?

Lol, NO, I'm not sure... probably got lost in the new rules!

Will send you my solution by PM in a minute.

Well.... oh dear... have to go and learn adding integers...

[moglet]

d_g, can you clarify d(ii)

Does this mean that the first colour cannot be potted into the the black or pink pockets, i.e. unless (or until) a previous colour.....

[abextra]

looks quite high abextra are you sure?

Lol, NO, I'm not sure... probably got lost in the new rules!

Will send you my solution by PM in a minute.

[davis_greatest]

Opening bid - 202 slaps. :snooker:

looks quite high abextra are you sure?

[abextra]

Opening bid - 202 slaps. :snooker:

[davis_greatest]

Round 365 - One Big Smack a Day

Big Smack is back, for a special Anniversary 365th edition!

The full, usual rules are pasted again below, with the special Anniversary edition updates in red:

============================================

On the show, contestants play a frame of snooker, just like any normal frame except that:

(a) Every point that a player scores permits him or her to give Damon Grott's face one nice, hearty slap.

(b) Each of the 6 pockets is coloured. The colours of the pockets are:
yellow for the left-centre pocket and then, moving clockwise, blue, brown, green, pink, black - a bit like this:

0------0
!.........!
!.........!
!.........!
0------0
!.........!
!.........!
!.........!
0------0

(c) Once a colour has been potted, the same colour cannot be potted following the next red, nor following the red after that. (Once the 15th red and colour have been potted, this rule no longer applies - the final colours must be potted in the usual order of yellow, green, brown, blue, pink, black, regardless of the colours potted with the final reds.)

Example 1: Red Brown Red Yellow Red Brown is NOT allowed; but Red Brown Red Yellow Red Blue Red Brown IS.

Example 2: For the 14th and 15th reds,
Red Black Red Yellow Yellow Green Brown Blue Pink Black IS allowed.

(d) Throughout the game (even when down to the final 6 colours), the following apply:

(i) A colour cannot be potted into the same pocket as the previous colour.
(ii) Any colour after the first cannot be potted into the pink or black pockets unless the previous colour was potted in the yellow pocket.
(iii) A colour cannot be potted into the blue pocket if the previous colour was potted in the green pocket.
(iv) A colour cannot be potted into the yellow pocket if the previous colour was potted in the brown or blue pockets.

(e) Rules (c) and (d) apply only to colours, not to reds. It makes no difference into which pockets reds are potted. Each red simply allows the player to give Damon one slap.

(f) Potting a colour into a pocket of the same colour as the ball scores double points (e.g. pink into pink pocket would be worth 12 slaps).

================================================== ===

You come to the table, with all 15 reds on the table after your opponent broke off (he did not foul). What is the greatest number of slaps, through judicious shot selection, you can give Damon during this frame?

You don't need to say the highest theoretically possible - just bid your slaps number directly here on the thread. Credit will be given not only for a large number of slaps, but also for speed.

[davis_greatest]

Fuller answer to round 364 in hidden text...

( Let's deal in units where the width of a snooker ball = 1.

The 50,100 specks are injected a distance at least 0.5 below the surface of the marble cube from every direction and can therefore be contained in a cube of dimensions 4x4x4 (call this the "Large Interior Cube").

Let's work out the width w of the largest cube (call it a "Mini Cube") that can fit inside a ball:

Applying Pythagoras, the square of the diagonal across the Mini Cube base will be w^2+w^2 = 2w^2. Applying Pythagoras again, the square of the long diagonal between opposite corners will be 2w^2+w^2 = 3w^2.

This long diagonal equals the diameter of a ball, so 3w^2 = 1^2 = 1, i.e. the Mini Cube has width w=1/sqrt 3.

The Large Interior Cube is 4 / w = 4 x sqrt(3) = approx 6.93 times as wide as a Mini Cube.

So let's cut the Large Interior Cube up into 7x7x7 = 343 smaller cubes - call these "Micro Cubes" (each with width 4/7 and slightly smaller than a Mini Cube).

The Micro Cubes therefore each contain on average 50,100 / 343 = approx 146.1 specks > 146 specks. Therefore, at least one chosen Micro Cube must contain at least 147 specks. (If none contained more than 146 specks, there would only be a maximum total of 343 x 146 = 50,078 specks, which isn't enough.)

We can then enclose that chosen Micro Cube by a Mini Cube, which in turn can be enclosed by a ball. (Since Charlie's sculpture has marble of width 0.5 around the Large Interior Cube, there is plenty of spare marble space to carve the parts of the ball outside the Micro (and Mini) Cube it contains.)
)

[davis_greatest]

...one can model the situation by centring those 146 specks in each cube ....

But how would one know that a cubic arrangement is the best way of spacing out as many clusters of specks as possible evenly (a ball's width apart)? One doesn't. It's a bit like the difficulty in packing in R361. You could try, for example, a triangular / pyramid lattice which may allow more clusters, still all more than a ball's width apart. We need to prove that what we have works in every case.

The answers already up can be combined to give a full solution, so I'll view this as collectively solved. Congratulations! I'll put up a fuller proof shortly...

[snookersfun]

OK what about this ...
Without the calculations just the idea
Let's divide the 4 X 4 X4 cube in smaller cubes taking the highest dimension being both smaller than the "ball inscribed" cube and allowing to divide exactly the 4x4x4 cube "side" into n (integer number n ).
We can then calculate the average of specks in one of those n^3 smaller cubes. If that is >147 ... it means that at least one of them contains at least 147 specks. As Charlie knows where the specks are he knows where this (or those) cubes are ... so he just can carve a ball "containing" it.

yes, that would prove it no doubt (of course >146, s.a.), just still think it is some kind of an overkill

Well... I'm really bad at explaining... ... if Gordon injected all the dust 146 spects per injection, he had to make at least 343 injections... and then the distances between the injections would be so small that a snooker ball would contain more than one of them... maybe I misunderstood the question again?

I still like this idea. Shouldn't this be considered a proof as well? As yes, one can divide the 4x4x4 volume into 343 subcubes, but one definitely can't fit 343 balls into the same volume (without trying to pack balls, I would assume not even half that amount).
Now, yes, those specks can be anywhere in each subcube, but in order to be as far away from EACH of the neighbouring volumes throughout the whole 4x4x4, one can model the situation by centring those 146 specks in each cube (actually could all be a bit off centre as the outside cubes could have their centre right on the outer surfaces. Sure one could do density plots as well, but that would result in the same general idea), thus one can carve a ball containing at least 2 of those centres.
Or looking at it from another angle, one would need to move those centres in neighbouring cubes at least 1 ballswidth apart, but can't do that throughout the whole grid.
I am sure one can express that brighter as well, but hope one gets the idea?

[davis_greatest]

OK what about this ...
Without the calculations just the idea
Let's divide the 4 X 4 X4 cube in smaller cubes taking the highest dimension being both smaller than the "ball inscribed" cube and allowing to divide exactly the 4x4x4 cube "side" into n (integer number n ).
We can then calculate the average of specks in one of those n^3 smaller cubes. If that is >147 ... it means that at least one of them contains at least 147 specks. As Charlie knows where the specks are he knows where this (or those) cubes are ... so he just can carve a ball "containing" it.

Yes! (apart from the >147 bit, which should say >146. Even 146.0001 (say) would do)

The main numbers needed are suggested in one of abextra's earlier posts.

[Monique]

OK what about this ...
Without the calculations just the idea
Let's divide the 4 X 4 X4 cube in smaller cubes taking the highest dimension being both smaller than the "ball inscribed" cube and allowing to divide exactly the 4x4x4 cube "side" into n (integer number n ).
We can then calculate the average of specks in one of those n^3 smaller cubes. If that is >147 ... it means that at least one of them contains at least 147 specks. As Charlie knows where the specks are he knows where this (or those) cubes are ... so he just can carve a ball "containing" it.

[davis_greatest]

But if they were NOT spread equally, then there were places in this 4x4x4 cube where there were more than 410 specks per ball?

That doesn't necessarily follow. It would follow that there were places where a shape with the same volume as a ball has more than 410 specks - but not necessarily that that shape would be the same shape as a ball!

[davis_greatest]

Yes but even if they are not spread equally ... IF no ball - wherever carved - contains more than 146 ... average density in the cube can't be 782,81 because that would require an average of about 407 specks per ball

A density argument on its own doesn't quite work. Here's an example why not:

Imagine we had only 4 specks injected and they were put at 4 opposite corners of a 1x1x1 cube - then any pair of specks is sqrt(2) apart.

Density would be 4 / Volume = 4/1 = 4 specks per unit volume.

Volume of a ball would be 4/3 x pi x (1/2)^3 = pi / 6 = 0.52 approx

So if we said average specks per ball were 4 x 0.52 = 2.08, that would suggest you could carve a ball with 2 specks at least, and indeed one with 3 specks!

But we know the specks are sqrt(2) = 1.4 approx ball-widths apart, so you can't even carve one ball with 2 specks.

Therefore we need to consider the shape, not just the volume of the ball.

[abextra]

Yes, there would be about 410 specks per ball IF the specks were spread evenly - but they are not necessarily.

But if they were NOT spread equally, then there were places in this 4x4x4 cube where there were more than 410 specks per ball?